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<article class="li"><h3 class="heading">
<span class="type">Item</span><span class="space"> </span><span class="codenumber">1</span><span class="period">.</span>
</h3>
<p>For the ODE problem</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2\frac{dy}{dt}-y=\sin t, \ \ y(0)=1. \label{ode1}
\end{equation*}
</div>
<ol class="lower-alpha">
<li>
<p>obtain the transformed version as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2[sY(s) -1]-Y(s) = \frac{1}{s^2+1}.
\end{equation*}
</div>
</li>
<li>
<p>Rearrange to get</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
Y(s)=\frac{2s^2+3}{(2s-1)(s^2+1)} =
\frac{A}{2s-1} +\frac{Bs+C}{s^2 +1}
\end{equation*}
</div>
</li>
<li>
<p>Show that <span class="process-math">\(A=\frac{14}{5}, \ B=\frac{-2}{5}, \ C=\frac{-1}{5},\)</span> and take the inverse transform to obtain the final solution to (<code class="code-inline tex2jax_ignore">[cross-reference to target(s) "ode1" missing or not unique]</code>) as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(t)= \frac{7}{5} e^{t/2} -\frac{2}{5} \cos t - \frac{1}{5}\sin t.
\end{equation*}
</div>
</li>
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